Problem: $f(x) = \begin{cases} \dfrac{1}{x} & \text{for} ~~~~x\gt{1} \\ x& \text{for} ~~~~ x \leq1\end{cases}$ Evaluate the definite integral. $\int^3_{0}f(x)\,dx = $ Choose 1 answer: Choose 1 answer: (Choice A) A $\ln(3) +\dfrac12$ (Choice B) B $\ln(\frac13) +1$ (Choice C) C $\ln(9) +\dfrac14$ (Choice D) D Undefined
Explanation: Splitting up the definite integral Since we're working with a piecewise function, we need to split the definite integral up into two pieces: $\phantom{=} \int^3_{0}f(x)\,dx$ $= \int^3_{1}f(x)\,dx + \int^{1}_{0}f(x)\,dx~~~~~~$ [Why did we split at 1?] $= \int^3_{1}\dfrac1x\,dx + \int^{1}_{0}x\,dx ~~~~~~$ Evaluating each piece Next, let's evaluate each of these definite integrals one at a time. The first definite integral: $\begin{aligned} \int^3_{1}\dfrac1x\,dx &=\ln(x)\Bigg|^3_{{1}} \\\\ &= \left[\ln( 3) \right] - \left[\ln ({1})\right] \\\\ &= \left[\ln(3)\right] -\left[0 \right] \\\\ &= {\ln(3)}\end{aligned}$ The second definite integral: $\begin{aligned} \int^{1}_{0}x\,dx &=\dfrac12x^2\Bigg|^1_{{0}} \\\\ &= \left[\dfrac12\cdot\left({1} \right)^2 \right] - \left[ ({0})^2\right] \\\\ &= \left[\dfrac12\right] -\left[0 \right] \\\\ &= {\dfrac12}\end{aligned}$ Putting the pieces together Now let's add these two pieces together to find the answer: $\phantom{=} \int^3_{1}\dfrac1x\,dx + \int^{1}_{0}x\,dx$ $ = {\ln(3)} + {\dfrac12}$ The answer $\int^3_{0}f(x)\,dx = \ln(3) +\dfrac12$